B. Sum of Two Numbers

B. Sum of Two Numbers

The sum of digits of a non-negative integer $a$ is the result of summing up its digits together when written in the decimal system. For example, the sum of digits of $123$ is $6$ and the sum of digits of $10$ is $1$. In a formal way, the sum of digits of $\displaystyle a=\sum_{i=0}^{\infty} a_i \cdot 10^i$, where $0 \leq a_i \leq 9$, is defined as $\displaystyle\sum_{i=0}^{\infty}{a_i}$.

Given an integer $n$, find two non-negative integers $x$ and $y$ which satisfy the following conditions.

• $x+y=n$, and
• the sum of digits of $x$ and the sum of digits of $y$ differ by at most $1$.

It can be shown that such $x$ and $y$ always exist.

Input

Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10\,000$).

Each test case consists of a single integer $n$ ($1 \leq n \leq 10^9$)

Output

For each test case, print two integers $x$ and $y$.

If there are multiple answers, print any.

Example

input

5
1
161
67
1206
19

output

1 0
67 94
60 7
1138 68
14 5

Note

In the second test case, the sum of digits of $67$ and the sum of digits of $94$ are both $13$.

In the third test case, the sum of digits of $60$ is $6$, and the sum of digits of $7$ is $7$.

 

解题思路

　　经典思维题想不出来。比赛的时候想了很久，最后还是写暴搜过的，先贴个代码。

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;

int n;
vector<int> num;
int pow10[10];
int ans[2];

int get(int x) {
    int ret = 0;
    while (x) {
        ret += x % 10;
        x /= 10;
    }
    return ret;
}

bool dfs(int a, int b, int u, int r) {
    if (u == num.size() && a + b == n) {
        ans[0] = a, ans[1] = b;
        return true;
    }
    for (int i = 0; i <= 9; i++) {
        int j = ((num[u] - i - r) % 10 + 10) % 10;
        int t = abs(get(i * pow10[u] + a) - get(j * pow10[u] + b));
        if (t <= 1 && dfs(i * pow10[u] + a, j * pow10[u] + b, u + 1, (i + j) / 10)) return true;
    }
    return false;
}

void solve() {
    scanf("%d", &n);
    num.clear();
    for (int i = n; i; i /= 10) {
        num.push_back(i % 10);
    }
    pow10[0] = 1;
    for (int i = 1; i < num.size(); i++) {
        pow10[i] = pow10[i - 1] * 10;
    }
    dfs(0, 0, 0, 0);
    printf("%d %d\n", ans[0], ans[1]);
}

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        solve();
    }
    
    return 0;
}

　　思路就是从低位开始往高位搜，假设$n$的第$i$位是$a_i$，$y$的第$i$位是$b_i$，$x$的第$i$位是$c_i$，然后从$0 \sim 9$枚举$b_i$，那么$c_i = (a_i - b_i - r) \bmod 10$，其中$r$是来自低位的进位，同时还要满足$b_i + c_i \leq s$，$s$是指允许两数相差的最大值。实际上加上这些约束条件的剪枝后有效状态是非常少的（大概十多个，其中只要看哪个位两数相差为$1$就好了），因此可以过。

　　实际上这是道思维题来的。每一位都可以看作是独立的，因此可以没有进位。如果$n$是偶数的话那么直接$x = y = \frac{n}{2}$就好了。否则$n$是奇数，枚举$n$的每一位，如果$a_i$是偶数，那么$b_i = c_i = \frac{a_i}{2}$。如果$a_i$是奇数，那么$b_i$和$c_i$一个取$\left\lfloor \frac{a_i}{2} \right\rfloor$，一个取$\left\lceil \frac{a_i}{2} \right\rceil$，为了保证最后所有位数之和相差不超过$1$，可以交替地给$b_i$和$c_i$分配较大的数和较小的数。

　　AC代码如下：

#include <bits/stdc++.h>
using namespace std;

void solve() {
    int n;
    scanf("%d", &n);
    int a = 0, b = 0, s = 0;
    for (auto &c : to_string(n)) {
        int x = c - '0';
        if (x & 1) {
            a = a * 10 + (x + s) / 2;   // s是奇数，那么把较大的数分配给a
            b = b * 10 + (x + !s) / 2;  // s是偶数，那么把较大的数分配给b
            s ^= 1;
        }
        else {
            a = a * 10 + x / 2;
            b = b * 10 + x / 2;
        }
    }
    printf("%d %d\n", a, b);
}

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        solve();
    }
    
    return 0;
}

 

参考资料

　　Codeforces Round #851 (Div. 2) Editorial：https://codeforces.com/blog/entry/112584